Question

A particle is thrown from the ground with u m/s upwards. It crosses the same height at 3 and 9s (9=10 m/s2)
Find the max height attained by particle from ground.

Answer 1

Answer:  I think so the answer must be 180m

              MARK ME AS BRAINLIEST!!!!!!!!

             hOPE THIS HELPS YOU :)

Explanation:  Let, height = h

                              t1 = 3 sec,     t2 = 9 sec

  h = ut1 + 1/2gt1²  = 3u + 1/2 * (-10) * 9  = 3u - 45    ----------I   { Acc= negative as it is acting in the opposite direction}

 h = ut2 + 1/2gt2²  = 9u + 1/2 * 10 * 81 = 9u + 405   ---------- II

Since height is same, therefore

Equating equations I and II, we will obtain u= -75m/s

Putting the value of u in I, we get h = 180m